Chapter 4. The First Fundamental Form (Induced Metric)

Chapter 4. The First Fundamental Form (Induced Metric) We begin with some definitions from linear algebra. Def. Let V be a vector space (over IR). A bilinear form on V is a map of the form B : V V IR which is bilinear, i.e. linear in each slot, B(aX + by, Z) ab(x, Z) + bb(y, Z), B(X, cy + dz) cb(x, Y ) + db(x, Z). A bilinear form B is symmetric provided B(X, Y ) B(Y, X) for all X, Y V. Def. Let V be a vector space. An inner product on V is a bilinear form, : V V IR which is symmetric and positive definite. Ex. 1. bilinear: linear in each slot, 2. symmetric: X, Y Y, X for all X, Y. 3. positive definite: X, X 0 X, and 0 iff X 0., : T p IR 3 T p IR 3 IR, Exercise 4.1 product. X, Y X Y (usual Euclidean dot product). Verify carefully that the Euclidean dot product is indeed an inner Def. Let M be a surface. A metric on M is an assignment, to each point p M, of an inner product, : T p M T p M IR. Because our surfaces sit in Euclidean space, they inherit in a natural way, a metric called the induced metric or first fundamental form. Def. Let M be a surface. The induced metric (or first fundamental form) of M is the assignment to each p M of the inner product,, : T p M T p M IR, X, Y X Y (ordinary scalar product of X and Y viewed as vectors in IR 3 at p) 1

I.e., the induced metric is just the Euclidean dot product, restricted to the tangent spaces of M. We will only consider surfaces in the induced metric. Just as the Euclidean dot product contains all geometric information about IR 3, the induced metric contains all geometric information about M, as we shall see. The Metric in a Coordinate Patch. Let x : U IR 2 M IR 3 be a proper patch in M. Let p x(u) be any point in x(u), p x(u 1, u 2 ), and let X, Y T p M. Then, X X 1 x x + X2 u1 u 2 X1 x 1 + X 2 x 2, X i X i x i, x i x i (u 1, u 2 ), and similarly, Then, Y j Y j x j. X, Y i X i x i, j Y j x j i,j X i Y j x i, x j. The metric components are the functions : U IR, 1 i, j 2, defined by x i, x j, (u 1, u 2 ). Thus, in coordinates, 2 X, Y X i Y j. i,j1 2

Note that the metric in x(u) is completely determined by the s. The metric components may be displayed by a 2 2 matrix, g 11 g 12 [ ], g 21 g 22 Note: x i, x j x j, x i g ji. Hence, the matrix of metric components is symmetric; and there are only three distinct components, g 11 x 1, x 1, g 12 x 1, x 2 x 2, x 1 g 21, g 22 x 2, x 2 Notation: 1. Gauss: g 11 E, g 12 g 21 F, g 22 G. 2. x(u, v) (x(u, v), y(u, v), z(u, v)). Then one writes: g uu x u, x u, g uv x u, x v, g vv x v, x v. Ex. Consider the parameterization of S 2 r in terms of geographic coordinates, x(θ, φ) (r sin θ cos φ, r sin θ sin φ, r cos θ), 0 < θ < π, 0 < φ < 2π. We compute the metric components in these coordinates. We have, x θ x θ r(cos θ cos φ, cos θ sin φ, sin θ), x φ r( sin θ sin φ, sin θ cos φ, 0), g θθ x θ, x θ r 2 [cos 2 θ cos 2 φ + cos 2 θ sin 2 φ + sin 2 θ] r 2 (cos 2 θ + sin 2 θ) r 2, g θφ x θ, x φ r 2 [ cos θ cos φ sin θ sin φ + cos θ sin φ sin θ cos φ] 0 (geometric significance?), g φφ r 2 [sin 2 θ sin 2 φ + sin 2 θ cos 2 φ] r 2 sin 2 θ. 3

Thus, [ gθθ g [ ] θφ g θφ g φφ ] [ r 2 0 0 r 2 sin 2 θ ] Length and Angle Measurement in M. Let σ : [a, b] M IR 3 be a smooth curve in a surface M. Viewed as a curve in IR 3, σ(t) (x(t), y(t), z(t)), we can compute its length by the formula, b Length of σ dσ a b (dx ) 2 ( ) 2 ( ) 2 dy dz + + a But this formula does not make sense to creatures living in the surface: x, y, z are Euclidean space coordinates. Creatures living in the surface must use surface coordinates i.e., we must express σ in terms of surface coordinates. Let x : U M IR 3 be a proper patch in M and suppose σ is contained in this patch, σ x(u) : We express σ in terms of coordinates: ˆσ x 1 σ : [a, b] U IR 2, ˆσ(t) (u (t), u 2 (t)), i.e, u ˆσ : 1 u 1 (t) u 2 u 2 (t), a t b. Then, σ x ˆσ, i.e., By the chain rule, σ(t) x(ˆσ(t)), hence, σ(t) x(u 1 (t), u 2 (t)). dσ x du 1 u 1 + x du 2 u 2 du1 x 1 + du2 x 2, 4

or, dσ i du i x i. This shows that dui, i 1, 2, are the components of the velocity vector with respect to the basis {x 1, x 2 }. Computing the dot product, dσ, dσ i du i x i, j du j x j i,j du i du j x i, x j 2 du i i,j1 du j. Hence, for the speed in surface coordinates, we have, dσ 2 du i du j. For length, we then have, Length of σ b a b a i,j du i g 11 ( du 1 i,j1 du j ) 2 + 2g 12 du 1 du 2 + g 22 ( ) du 2 2 Arc length. Let s denote arc length along σ, s can be computed in terms of t as follows. s s(t), a t b, s(t) length of σ from time a to time t t a i,j du i du j. 5

Arc length element: ds i,j du i du j In terms of differentials, ds ds 2 i,j ( i,j i,j du i du i du j ) du j 2 ( ) ( ) du i du j ds 2 Heuristics: element of arc length 2 du i du j. i,j1 ds 2 i,j du i du j Traditionally, one displays the metric (or, metric components ) by writing out the arc length element. Notations: ds 2 g 11 (du 1 ) 2 + 2g 12 du 1 du 2 + g 22 (du 2 ) 2 ds 2 g uu du 2 + 2g uv dudv + g vv dv 2 (u 1 u, u 2 v) ds 2 Edu 2 + 2F dudv + Gdv 2 (Gauss). 6

Remark. These expressions for arc length element of a surface M generalize the expression for the arc length element in the Euclidean u-v plane we encounter in calculus, ds 2 du 2 + dv 2 (i.e. g uu 1, g uv 0, g vv 1). Ex. Write out the arc length element for the sphere Sr 2 parameterized in terms of geographic coordinates, x(θ, φ) (r sin θ cos φ, r sin θ sin φ, r cos θ). We previously computed the s, [ gθθ g [ ] θφ g φθ g φφ ] [ r 2 0 0 r 2 sin 2 θ ] i.e., g θθ r 2, g θφ g φθ 0, g φφ r 2 sin 2 θ. So, ds 2 g θθ dθ 2 + 2g θφ dθdφ + g φφ dφ 2 ds 2 r 2 dθ 2 + r 2 sin 2 θdφ 2. But this expression is familiar from calculus as the arc length element which can be derived from heuristic geometric considerations. ds 2 dl 2 1 + dl 2 2. dl 1 rdθ, dl 2 r sin θdφ ds 2 r 2 dθ 2 + r 2 sin 2 θdφ 2. Exercise 4.2. coordinates, Consider the parameterization of the x-y plane in terms of polar x r cos θ x : y r sin θ, 0 < r <, 0 < θ < 2π, z 0 i.e., x(r, θ) (r cos θ, r sin θ, 0), 0 < r <, 0 < θ < 2π. Compute the s with respect to these coordinates. Show that the arc length element in this case is: ds 2 dr 2 + r 2 dθ 2. 7

Angle Measurement. X i Y j X i x i, Y j x j, cos θ X, Y X Y gij X i Y j gij X i X j gij Y i Y j. Ex. Determine the angle between the coordinate vectors x 1 x u and x 1 2 x u 2 in terms of the s. cos θ x 1, x 2 x 1 x 2 g 12 g11 g22 ( x 1 x 1, x 1 g 11, etc.) The Metric is intrinsic: This discussion is somewhat heuristic. We claim that the s are intrinsic, i.e. in principle they can be determined by measurements made in the surface. Let x : U M be a proper patch in M; x x(u 1, u 2 ) x(u, v) (i.e., u 1 u, u 2 v. Consider the coordinate curve u σ x(u, v 0 ) passing through x(u 0, v 0 ). 8

Let s s(u) be the arc length function along σ, i.e., s(u) length of σ from u 0 to u u u 0 u x u du u 0 guu du ( x u ) g uu. By making length measurements in the surface the function s s(u) is known. Then by calculus, the derivative, ds du g uu. is known. Therefore g 11 g uu, and similarly g 22 g vv, can in principal be determined by measurements made in the surface. The metric component g 12 can then be determined by angle measurement, g 12 x 1, x 2 x 1 x 2 cos θ g 11 g22 cos(angle between x 1, x 2 ). Hence g 12 is also measurable. Thus all metric components can be determined by measurements made in the surface, i.e. the metric components and all quantities determined from them are intrinsic. Surface Area. Let M be a surface, and let x : U M be a proper patch in M. Consider a bounded region R contained in x(u); we have R x(w) for some bounded region W in U: We want to obtain (i.e. heuristically motivate) a formula for the area of R x(w ). Restrict attention to R x(w ); partition W into small rectangles: 9

Let S area of the small patch corresponding to the coordinate rectangle. Then, S area of the parallelogram spanned by P Q and S P Q P R. P R, But, and thus, P Q x(u + u, v) x(u, v) x u u, P R x(u, v + v) x(u, v) x v v, S x x u u v v x u x v u v. The smaller the increments u and v, the better the approximation. ds the area element of the surface corresponding to the ds coordinate increments du, dv, x u x v du dv. To obtain the total area of R, we must sum up all these area elements - but the summing up process is integration: Area of R ds, Area of R x u v du dv, W 10

where R x(w ). This is a perfectly reasonable formula for computing surface area - but not for 2-dimensional creatures living in the surface. It involves the cross product which is an IR 3 concept. We now show how this area formula can be expressed in an intrinsic way (i.e. involving the s ). Using generic notation, u 1 u, u 2 v, x x(u 1, u 2 ) we write, Area of R x u x 1 u 2 du1 du 2 Now introduce the notation, Lemma. g x 1 x 2 2 W W x 1 x 2 du 1 du 2 g det[ ], x i, x j. Proof. Hence, Recall the vector identity, a b 2 a 2 b 2 (a b) 2. x 1 x 2 2 x 1 2 x 2 2 x 1, x 2 2 x 1, x 1 x 2, x 2 x 1, x 2 2 g 11 g 22 g12 2 g, g 11 g 12 g det[ ] det g 11 g 22 g12 2, g 21 g 22 where we have used g 21 g 12. Thus, the surface area formula may be expressed as, where, Area of R g du 1 du 2 (R x(w )) W ds, W ds g du 1 du 2. 11

Ex. Compute the area of the sphere of radius r. S 2 r : x 2 + y 2 + z 2 r 2. Parameterize with respect to geographical coordinates, x : U S 2 r, U : 0 < θ < π, 0 < φ < 2π. x(θ, φ) (r sin θ cos φ, r sin θ sin φ, r cos θ), Now, Thus, We have, Area of Sr 2 g g ds U ds, where ds gdθdφ. [ gθθ g det[ ] det θφ g φθ g φφ [ r 2 0 det 0 r 2 sin 2 θ r 4 sin 2 θ r 4 sin 2 θ dθdφ r 2 sin θdθdφ ] ] Remark: This expression for the surface area element of a sphere is familiar from calculus or physics where it is usually derived by heuristic considerations: 12

ds dl 1 dl 2, dl 1 ds rdθ, dl 2 r sin θdφ (rdθ)(r sin θdφ) r 2 sin θdθdφ. Continuing the computation of the surface area of Sr 2, Area of Sr 2 r 2 sin θdθdφ r 2 sin θdθdφ U 2π π 0 2π 0 0 r 2 sin θdθdφ U 2π 2r 2 dφ 2r 2 φ 2π 0 4πr 2. 0 r 2 [ cos θ] π 0 dφ The surface area formula involves a choice of coordinates, i.e. a choice of proper patch. It is important to recognize that the formula is independent of this choice. Proposition. The area formula is independent of the choice of coordinate patch. Let x : U M, x(u) y(v ): y : V M be proper patches, and suppose R is contained in Set, x i, x j, g det[ ], y i, y j, g det[ ]. 13

Then the claim is that, x 1 (R) gdu 1 du 2 y 1 (R) gdv 1 dv 2 Proof: The proof is an application of the change of variable formula for double integrals. Let f : U IR 2 V IR 2 be a diffeomorphism, where U, V are bounded regions in IR 2. f : v 1 f 1 (u 1, u 2 ) v 2 f 2 (u 1, u 2 ). Then, the change of variable formula for double integrals is as follows, h(v 1, v 2 )dv 1 dv 2 h f(u 1, u 2 ) det Df du 1 du 2 V or, in briefer notation, V U U hdv 1 dv 2 h(f 1 (u 1, u 2 ), f 1 (u 1, u 2 )) (v 1, v 2 ) (u 1, u 2 ) du1 du 2 U h (v 1, v 2 ) (u 1, u 2 ) du1 du 2., In the case at hand, f y 1 x : x 1 (R) y 1 (R), and h g. So, by the change of variable formula, gdv 1 dv 2 (v g 1, v 2 ) (u 1, u 2 ) du1 du 2 y 1 (R) x 1 (R) Thus, to complete the proof, it suffices to establish the following lemma. Lemma. g det[ ], g det[ ]. are related by, (v g g 1, v 2 ) (u 1, u 2 ). 14

Proof of the lemma: By the Exercise 3.4 (but with role of x and y reversed), x u x 1 u (v1, v 2 ) 2 (u 1, u 2 ) y v 1 y dv 2 or, x 1 x 2 (v1, v 2 ) (u 1, u 2 ) y 1 y 2 Hence, Taking square roots yields the result. g det[ ] x 1 x 2 2 ( ) (v 1, v 2 2 ) y (u 1, u 2 1 y 2 2 ) ( ) (v 1, v 2 2 ) g. (u 1, u 2 ) Exercise 4.3 Consider the torus of large radius R and small radius r described in Exercise 3.3. Use the intrinsic surface area formula and the parameterization given in Exercise 3.3 to compute the surface area of the torus. Answer: 4π 2 Rr. Exercise 4.4 Let f : U IR 2 IR be a smooth function of two variables. Let M be the graph of f w {(x, y, z) IR 3 : z f(x, y), (x, y) W }, where W is a bounded subset of U. Derive the following standard formula from calculus for the surface area of M, ( ) 2 ( ) 2 f f Area of M 1 + + dxdy, x y by considering the Monge patch associated to f. More Tensor Analysis W Let x : U M, y : V M be overlapping patches in a surface M, W : x(u) y(v ). Let f y 1 x : x 1 (W ) y 1 (W ), f : v 1 f 1 (u 1, u 2 ) v 2 f 2 (u 1, u 2 ) 15

be the smooth overlap map, cf., p. 15 of Chapter 3. Introduce the metric components with respect to each patch, x i, x j, y i, y j. How are these metric components related on the overlap? Exercise 4.5 Show that, 2 v a v b g ab, i, j 1, 2. u i uj a,b1 These equations can be expressed as a single matrix equation, Taking determinants we obtain, [ ] v a t [ v b [ ] [ g ab] u i u j ]. g det[ ] det[ ] t [ ][ ] g det[ ] t det[ ] det[ ] [ ] [ v a v b det det[ g u i ij ] det u j g(det Df) 2 g our second derivation of this formula. [ ] (v 1, v 2 2 ), (u 1, u 2 ) ] Remark: Interchanging the roles of x and y above we obtain, g ab i,j u i v a u j v b, which involves the Jacobian of f 1. Compare this transformation law for the metric components to the transformation law for vector components considered in Exercise 3.8. Vector fields are contravariant tensors. The metric, is a covariant tensor. 16